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# Introduction to Disjoint Set (Union-Find Algorithm)

## What is a Disjoint set data structure?

Two sets are called disjoint sets if they don’t have any element in common, the intersection of sets is a null set.

A data structure that stores non overlapping or disjoint subset of elements is called disjoint set data structure. The disjoint set data structure supports following operations:

• Adding new sets to the disjoint set.
• Merging disjoint sets to a single disjoint set using Union operation.
• Finding representative of a disjoint set using Find operation.
• Check if two sets are disjoint or not.

Consider a situation with a number of persons and the following tasks to be performed on them:

• Add a new friendship relation, i.e. a person x becomes the friend of another person y i.e adding new element to a set.
• Find whether individual x is a friend of individual y (direct or indirect friend)

Examples:

We are given 10 individuals say, a, b, c, d, e, f, g, h, i, j

Following are relationships to be added:
a <-> b
b <-> d
c <-> f
c <-> i
j <-> e
g <-> j

Given queries like whether a is a friend of d or not. We basically need to create following 4 groups and maintain a quickly accessible connection among group items:
G1 = {a, b, d}
G2 = {c, f, i}
G3 = {e, g, j}
G4 = {h}

### Find whether x and y belong to the same group or not, i.e. to find if x and y are direct/indirect friends.

Partitioning the individuals into different sets according to the groups in which they fall. This method is known as a Disjoint set Union which maintains a collection of Disjoint sets and each set is represented by one of its members.

To answer the above question two key points to be considered are:

• How to Resolve sets? Initially, all elements belong to different sets. After working on the given relations, we select a member as a representative. There can be many ways to select a representative, a simple one is to select with the biggest index.
• Check if 2 persons are in the same group? If representatives of two individuals are the same, then they’ll become friends.

### Data Structures used are:

Array: An array of integers is called Parent[]. If we are dealing with N items, i’th element of the array represents the i’th item. More precisely, the i’th element of the Parent[] array is the parent of the i’th item. These relationships create one or more virtual trees.

Tree: It is a Disjoint set. If two elements are in the same tree, then they are in the same Disjoint set. The root node (or the topmost node) of each tree is called the representative of the set. There is always a single unique representative of each set. A simple rule to identify a representative is if ‘i’ is the representative of a set, then Parent[i] = i. If i is not the representative of his set, then it can be found by traveling up the tree until we find the representative.

## Operations on Disjoint Set Data Structures:

1. Find
2. Union

### 1. Find:

Can be implemented by recursively traversing the parent array until we hit a node that is the parent of itself.

## C++

 `// Finds the representative of the set``// that i is an element of` `#include``using` `namespace` `std;` `int` `find(``int` `i)` `{` `    ``// If i is the parent of itself``    ``if` `(parent[i] == i) {` `        ``// Then i is the representative of``        ``// this set``        ``return` `i;``    ``}``    ``else` `{` `        ``// Else if i is not the parent of``        ``// itself, then i is not the``        ``// representative of his set. So we``        ``// recursively call Find on its parent``        ``return` `find(parent[i]);``    ``}``}` `// The code is contributed by Nidhi goel`

## Java

 `// Finds the representative of the set``// that i is an element of``import` `java.io.*;` `class` `GFG {` `    ``static` `int` `find(``int` `i)` `    ``{` `        ``// If i is the parent of itself``        ``if` `(parent[i] == i) {` `            ``// Then i is the representative of``            ``// this set``            ``return` `i;``        ``}``        ``else` `{` `            ``// Else if i is not the parent of``            ``// itself, then i is not the``            ``// representative of his set. So we``            ``// recursively call Find on its parent``            ``return` `find(parent[i]);``        ``}``    ``}``}` `// The code is contributed by Nidhi goel`

## Python3

 `# Finds the representative of the set``# that i is an element of` `def` `find(i):` `    ``# If i is the parent of itself``    ``if` `(parent[i] ``=``=` `i):` `        ``# Then i is the representative of``        ``# this set``        ``return` `i``    ``else``:` `        ``# Else if i is not the parent of``        ``# itself, then i is not the``        ``# representative of his set. So we``        ``# recursively call Find on its parent``        ``return` `find(parent[i])` ` ``# The code is contributed by Nidhi goel`

## C#

 `using` `System;` `public` `class` `GFG{` `    ``// Finds the representative of the set``    ``// that i is an element of``    ``public` `static` `int` `find(``int` `i)``    ``{` `        ``// If i is the parent of itself``        ``if` `(parent[i] == i) {` `            ``// Then i is the representative of``            ``// this set``            ``return` `i;``        ``}``        ``else` `{` `            ``// Else if i is not the parent of``            ``// itself, then i is not the``            ``// representative of his set. So we``            ``// recursively call Find on its parent``            ``return` `find(parent[i]);``        ``}``    ``}``}`

## Javascript

 ``

Time complexity: This approach is inefficient and can take O(n) time in worst case.

### 2. Union:

It takes two elements as input and finds the representatives of their sets using the Find operation, and finally puts either one of the trees (representing the set) under the root node of the other tree.

## C++

 `// Unites the set that includes i``// and the set that includes j` `#include ``using` `namespace` `std;` `void` `union``(``int` `i, ``int` `j) {` `    ``// Find the representatives``    ``// (or the root nodes) for the set``    ``// that includes i``    ``int` `irep = ``this``.Find(i),` `    ``// And do the same for the set``    ``// that includes j``    ``int` `jrep = ``this``.Find(j);` `    ``// Make the parent of i’s representative``    ``// be j’s  representative effectively``    ``// moving all of i’s set into j’s set)``    ``this``.Parent[irep] = jrep;``}`

## Python3

 `# Unites the set that includes i``# and the set that includes j` `def` `union(parent, rank, i, j):``    ``# Find the representatives``    ``# (or the root nodes) for the set``    ``# that includes i``    ``irep ``=` `find(parent, i)``    ` `    ``# And do the same for the set``    ``# that includes j``    ``jrep ``=` `find(parent, j)``    ` `    ``# Make the parent of i’s representative``    ``# be j’s  representative effectively``    ``# moving all of i’s set into j’s set)``    ` `    ``parent[irep] ``=` `jrep`

## Javascript

 `// JavaScript code for the approach` `// Unites the set that includes i``// and the set that includes j``function` `union(parent, rank, i, j)``{` `// Find the representatives``// (or the root nodes) for the set``// that includes i``let irep = find(parent, i);` `// And do the same for the set``// that includes j``let jrep = find(parent, j);` `// Make the parent of i’s representative``// be j’s representative effectively``// moving all of i’s set into j’s set)` `parent[irep] = jrep;``}`

Time complexity: This approach is inefficient and could lead to tree of length O(n) in worst case.

### Optimizations (Union by Rank/Size and Path Compression):

The efficiency depends heavily on which tree get attached to the other. There are 2 ways in which it can be done. First is Union by Rank, which considers height of the tree as the factor and Second is Union by Size, which considers size of the tree as the factor while attaching one tree to the other . This method along with Path Compression gives complexity of nearly constant time.

### Path Compression (Modifications to Find()):

It speeds up the data structure by compressing the height of the trees. It can be achieved by inserting a small caching mechanism into the Find operation. Take a look at the code for more details:

## C++

 `// Finds the representative of the set that i``// is an element of.` `#include ``using` `namespace` `std;` `int` `find(``int` `i)``{` `    ``// If i is the parent of itself``    ``if` `(Parent[i] == i) {` `        ``// Then i is the representative``        ``return` `i;``    ``}``    ``else` `{` `        ``// Recursively find the representative.``        ``int` `result = find(Parent[i]);` `        ``// We cache the result by moving i’s node``        ``// directly under the representative of this``        ``// set``        ``Parent[i] = result;``      ` `        ``// And then we return the result``        ``return` `result;``     ``}``}`

## Java

 `// Finds the representative of the set that i``// is an element of.``import` `java.io.*;``import` `java.util.*;` `static` `int` `find(``int` `i)``{` `    ``// If i is the parent of itself``    ``if` `(Parent[i] == i) {` `        ``// Then i is the representative``        ``return` `i;``    ``}``    ``else` `{` `        ``// Recursively find the representative.``        ``int` `result = find(Parent[i]);` `        ``// We cache the result by moving i’s node``        ``// directly under the representative of this``        ``// set``        ``Parent[i] = result;``      ` `        ``// And then we return the result``        ``return` `result;``     ``}``}` `// The code  is contributed by Arushi jindal.`

## Python3

 `#  Finds the representative of the set that i``# is an element of.`  `def` `find(i):` `    ``# If i is the parent of itself``    ``if` `Parent[i] ``=``=` `i:` `        ``# Then i is the representative``        ``return` `i``    ``else``:` `        ``# Recursively find the representative.``        ``result ``=` `find(Parent[i])` `        ``# We cache the result by moving i’s node``        ``# directly under the representative of this``        ``# set``        ``Parent[i] ``=` `result``      ` `        ``# And then we return the result``        ``return` `result` `# The code is contributed by Arushi  Jindal.`

## C#

 `using` `System;` `// Finds the representative of the set that i``// is an element of.``public` `static` `int` `find(``int` `i)``{` `    ``// If i is the parent of itself``    ``if` `(Parent[i] == i) {` `        ``// Then i is the representative``        ``return` `i;``    ``}``    ``else` `{` `        ``// Recursively find the representative.``        ``int` `result = find(Parent[i]);` `        ``// We cache the result by moving i’s node``        ``// directly under the representative of this``        ``// set``        ``Parent[i] = result;``      ` `        ``// And then we return the result``        ``return` `result;``     ``}``}` `// The code is contributed by Arushi Jindal.`

## Javascript

 `// Finds the representative of the set that i``// is an element of.`  `function` `find(i)``{` `    ``// If i is the parent of itself``    ``if` `(Parent[i] == i) {` `        ``// Then i is the representative``        ``return` `i;``    ``}``    ``else` `{` `        ``// Recursively find the representative.``        ``let result = find(Parent[i]);` `        ``// We cache the result by moving i’s node``        ``// directly under the representative of this``        ``// set``        ``Parent[i] = result;``      ` `        ``// And then we return the result``        ``return` `result;``     ``}``}` `// The code is contributed by Arushi  Jindal.`

Time Complexity: O(log n) on average per call.

### Union by Rank:

First of all, we need a new array of integers called rank[]. The size of this array is the same as the parent array Parent[]. If i is a representative of a set, rank[i] is the height of the tree representing the set.
Now recall that in the Union operation, it doesn’t matter which of the two trees is moved under the other. Now what we want to do is minimize the height of the resulting tree. If we are uniting two trees (or sets), let’s call them left and right, then it all depends on the rank of left and the rank of right

• If the rank of left is less than the rank of right, then it’s best to move left under right, because that won’t change the rank of right (while moving right under left would increase the height). In the same way, if the rank of right is less than the rank of left, then we should move right under left.
• If the ranks are equal, it doesn’t matter which tree goes under the other, but the rank of the result will always be one greater than the rank of the trees.

## C++

 `// Unites the set that includes i and the set``// that includes j by rank` `#include ``using` `namespace` `std;` `void` `unionbyrank(``int` `i, ``int` `j) {` `    ``// Find the representatives (or the root nodes)``    ``// for the set that includes i``    ``int` `irep = ``this``.find(i);` `    ``// And do the same for the set that includes j``    ``int` `jrep = ``this``.Find(j);` `    ``// Elements are in same set, no need to``    ``// unite anything.``    ``if` `(irep == jrep)``        ``return``;``    ` `      ``// Get the rank of i’s tree``    ``irank = Rank[irep],` `    ``// Get the rank of j’s tree``    ``jrank = Rank[jrep];` `    ``// If i’s rank is less than j’s rank``    ``if` `(irank < jrank) {` `        ``// Then move i under j``        ``this``.parent[irep] = jrep;``    ``}` `    ``// Else if j’s rank is less than i’s rank``    ``else` `if` `(jrank < irank) {` `        ``// Then move j under i``        ``this``.Parent[jrep] = irep;``    ``}` `    ``// Else if their ranks are the same``    ``else` `{` `        ``// Then move i under j (doesn’t matter``        ``// which one goes where)``        ``this``.Parent[irep] = jrep;` `        ``// And increment the result tree’s``        ``// rank by 1``        ``Rank[jrep]++;``    ``}``}`

## Javascript

 `// JavaScript Program for the above approach``unionbyrank(i, j) {``let irep = ``this``.find(i); ``// Find representative of set including i``let jrep = ``this``.find(j); ``// Find representative of set including j` `if` `(irep === jrep) {``return``; ``// Elements are already in the same set``}` `let irank = ``this``.rank[irep]; ``// Rank of set including i``let jrank = ``this``.rank[jrep]; ``// Rank of set including j` `if` `(irank < jrank) {``this``.parent[irep] = jrep; ``// Make j's representative parent of i's representative``} ``else` `if` `(jrank < irank) {``this``.parent[jrep] = irep; ``// Make i's representative parent of j's representative``} ``else` `{``this``.parent[irep] = jrep; ``// Make j's representative parent of i's representative``this``.rank[jrep]++;        ``// Increment the rank of the resulting set``}`

### Union by Size:

Again, we need a new array of integers called size[]. The size of this array is the same as the parent array Parent[]. If i is a representative of a set, size[i] is the number of the elements in the tree representing the set.
Now we are uniting two trees (or sets), let’s call them left and right, then in this case it all depends on the size of left and the size of right tree (or set).

• If the size of left is less than the size of right, then it’s best to move left under right and increase size of right by size of left. In the same way, if the size of right is less than the size of left, then we should move right under left. and increase size of left by size of right.
• If the sizes are equal, it doesn’t matter which tree goes under the other.

## C++

 `// Unites the set that includes i and the set``// that includes j by size` `#include ``using` `namespace` `std;` `void` `unionbysize(``int` `i, ``int` `j) {` `    ``// Find the representatives (or the root nodes)``    ``// for the set that includes i``    ``int` `irep = ``this``.find(i);` `    ``// And do the same for the set that includes j``    ``int` `jrep = ``this``.Find(j);` `    ``// Elements are in same set, no need to``    ``// unite anything.``    ``if` `(irep == jrep)``        ``return``;``    ` `      ``// Get the size of i’s tree``    ``isize = Size[irep],` `    ``// Get the size of j’s tree``    ``jsize = Size[jrep];` `    ``// If i’s size is less than j’s size``    ``if` `(isize < jsize) {` `        ``// Then move i under j``        ``this``.parent[irep] = jrep;``      ` `      ``// Increment j's size by i'size``        ``Size[jrep]+=Size[irep];``    ``}` `    ``// Else if j’s rank is less than i’s rank``    ``else` `if` `(jsize < isize) {` `        ``// Then move j under i``        ``this``.Parent[jrep] = irep;``      ` `      ``// Increment i's size by j'size``        ``Size[irep]+=Size[jrep];``    ``}` `    ``// Else if their ranks are the same``    ``else` `{` `        ``// Then move i under j (doesn’t matter``        ``// which one goes where)``        ``this``.Parent[irep] = jrep;` `        ``// Increment j's size by i'size``        ``Size[jrep]+=Size[irep];``    ``}``}`

Time complexity: O(log n) without Path Compression.

## C++

 `// C++ implementation of disjoint set` `#include ``using` `namespace` `std;` `class` `DisjSet {``    ``int` `*rank, *parent, n;` `public``:``  ` `    ``// Constructor to create and``    ``// initialize sets of n items``    ``DisjSet(``int` `n)``    ``{``        ``rank = ``new` `int``[n];``        ``parent = ``new` `int``[n];``        ``this``->n = n;``        ``makeSet();``    ``}` `    ``// Creates n single item sets``    ``void` `makeSet()``    ``{``        ``for` `(``int` `i = 0; i < n; i++) {``            ``parent[i] = i;``        ``}``    ``}` `    ``// Finds set of given item x``    ``int` `find(``int` `x)``    ``{``        ``// Finds the representative of the set``        ``// that x is an element of``        ``if` `(parent[x] != x) {` `            ``// if x is not the parent of itself``            ``// Then x is not the representative of``            ``// his set,``            ``parent[x] = find(parent[x]);` `            ``// so we recursively call Find on its parent``            ``// and move i's node directly under the``            ``// representative of this set``        ``}` `        ``return` `parent[x];``    ``}` `    ``// Do union of two sets by rank represented``    ``// by x and y.``    ``void` `Union(``int` `x, ``int` `y)``    ``{``        ``// Find current sets of x and y``        ``int` `xset = find(x);``        ``int` `yset = find(y);` `        ``// If they are already in same set``        ``if` `(xset == yset)``            ``return``;` `        ``// Put smaller ranked item under``        ``// bigger ranked item if ranks are``        ``// different``        ``if` `(rank[xset] < rank[yset]) {``            ``parent[xset] = yset;``        ``}``        ``else` `if` `(rank[xset] > rank[yset]) {``            ``parent[yset] = xset;``        ``}` `        ``// If ranks are same, then increment``        ``// rank.``        ``else` `{``            ``parent[yset] = xset;``            ``rank[xset] = rank[xset] + 1;``        ``}``    ``}``};` `// Driver Code``int` `main()``{``  ` `      ``// Function Call``    ``DisjSet obj(5);``    ``obj.Union(0, 2);``    ``obj.Union(4, 2);``    ``obj.Union(3, 1);``  ` `    ``if` `(obj.find(4) == obj.find(0))``        ``cout << ``"Yes\n"``;``    ``else``        ``cout << ``"No\n"``;``    ``if` `(obj.find(1) == obj.find(0))``        ``cout << ``"Yes\n"``;``    ``else``        ``cout << ``"No\n"``;` `    ``return` `0;``}`

## Java

 `// A Java program to implement Disjoint Set Data``// Structure.``import` `java.io.*;``import` `java.util.*;` `class` `DisjointUnionSets {``    ``int``[] rank, parent;``    ``int` `n;` `    ``// Constructor``    ``public` `DisjointUnionSets(``int` `n)``    ``{``        ``rank = ``new` `int``[n];``        ``parent = ``new` `int``[n];``        ``this``.n = n;``        ``makeSet();``    ``}` `    ``// Creates n sets with single item in each``    ``void` `makeSet()``    ``{``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``// Initially, all elements are in``            ``// their own set.``            ``parent[i] = i;``        ``}``    ``}` `    ``// Returns representative of x's set``    ``int` `find(``int` `x)``    ``{``        ``// Finds the representative of the set``        ``// that x is an element of``        ``if` `(parent[x] != x) {``            ``// if x is not the parent of itself``            ``// Then x is not the representative of``            ``// his set,``            ``parent[x] = find(parent[x]);` `            ``// so we recursively call Find on its parent``            ``// and move i's node directly under the``            ``// representative of this set``        ``}` `        ``return` `parent[x];``    ``}` `    ``// Unites the set that includes x and the set``    ``// that includes x``    ``void` `union(``int` `x, ``int` `y)``    ``{``        ``// Find representatives of two sets``        ``int` `xRoot = find(x), yRoot = find(y);` `        ``// Elements are in the same set, no need``        ``// to unite anything.``        ``if` `(xRoot == yRoot)``            ``return``;` `        ``// If x's rank is less than y's rank``        ``if` `(rank[xRoot] < rank[yRoot])` `            ``// Then move x under y  so that depth``            ``// of tree remains less``            ``parent[xRoot] = yRoot;` `        ``// Else if y's rank is less than x's rank``        ``else` `if` `(rank[yRoot] < rank[xRoot])` `            ``// Then move y under x so that depth of``            ``// tree remains less``            ``parent[yRoot] = xRoot;` `        ``else` `// if ranks are the same``        ``{``            ``// Then move y under x (doesn't matter``            ``// which one goes where)``            ``parent[yRoot] = xRoot;` `            ``// And increment the result tree's``            ``// rank by 1``            ``rank[xRoot] = rank[xRoot] + ``1``;``        ``}``    ``}``}` `// Driver code``public` `class` `Main {``    ``public` `static` `void` `main(String[] args)``    ``{``        ``// Let there be 5 persons with ids as``        ``// 0, 1, 2, 3 and 4``        ``int` `n = ``5``;``        ``DisjointUnionSets dus =``                ``new` `DisjointUnionSets(n);` `        ``// 0 is a friend of 2``        ``dus.union(``0``, ``2``);` `        ``// 4 is a friend of 2``        ``dus.union(``4``, ``2``);` `        ``// 3 is a friend of 1``        ``dus.union(``3``, ``1``);` `        ``// Check if 4 is a friend of 0``        ``if` `(dus.find(``4``) == dus.find(``0``))``            ``System.out.println(``"Yes"``);``        ``else``            ``System.out.println(``"No"``);` `        ``// Check if 1 is a friend of 0``        ``if` `(dus.find(``1``) == dus.find(``0``))``            ``System.out.println(``"Yes"``);``        ``else``            ``System.out.println(``"No"``);``    ``}``}`

## Python3

 `# Python3 program to implement Disjoint Set Data``# Structure.` `class` `DisjSet:``    ``def` `__init__(``self``, n):``        ``# Constructor to create and``        ``# initialize sets of n items``        ``self``.rank ``=` `[``1``] ``*` `n``        ``self``.parent ``=` `[i ``for` `i ``in` `range``(n)]`  `    ``# Finds set of given item x``    ``def` `find(``self``, x):``        ` `        ``# Finds the representative of the set``        ``# that x is an element of``        ``if` `(``self``.parent[x] !``=` `x):``            ` `            ``# if x is not the parent of itself``            ``# Then x is not the representative of``            ``# its set,``            ``self``.parent[x] ``=` `self``.find(``self``.parent[x])``            ` `            ``# so we recursively call Find on its parent``            ``# and move i's node directly under the``            ``# representative of this set` `        ``return` `self``.parent[x]`  `    ``# Do union of two sets represented``    ``# by x and y.``    ``def` `Union(``self``, x, y):``        ` `        ``# Find current sets of x and y``        ``xset ``=` `self``.find(x)``        ``yset ``=` `self``.find(y)` `        ``# If they are already in same set``        ``if` `xset ``=``=` `yset:``            ``return` `        ``# Put smaller ranked item under``        ``# bigger ranked item if ranks are``        ``# different``        ``if` `self``.rank[xset] < ``self``.rank[yset]:``            ``self``.parent[xset] ``=` `yset` `        ``elif` `self``.rank[xset] > ``self``.rank[yset]:``            ``self``.parent[yset] ``=` `xset` `        ``# If ranks are same, then move y under``        ``# x (doesn't matter which one goes where)``        ``# and increment rank of x's tree``        ``else``:``            ``self``.parent[yset] ``=` `xset``            ``self``.rank[xset] ``=` `self``.rank[xset] ``+` `1` `# Driver code``obj ``=` `DisjSet(``5``)``obj.Union(``0``, ``2``)``obj.Union(``4``, ``2``)``obj.Union(``3``, ``1``)``if` `obj.find(``4``) ``=``=` `obj.find(``0``):``    ``print``(``'Yes'``)``else``:``    ``print``(``'No'``)``if` `obj.find(``1``) ``=``=` `obj.find(``0``):``    ``print``(``'Yes'``)``else``:``    ``print``(``'No'``)` `# This code is contributed by ng24_7.`

## C#

 `// A C# program to implement ``// Disjoint Set Data Structure.``using` `System;``    ` `class` `DisjointUnionSets``{``    ``int``[] rank, parent;``    ``int` `n;` `    ``// Constructor``    ``public` `DisjointUnionSets(``int` `n)``    ``{``        ``rank = ``new` `int``[n];``        ``parent = ``new` `int``[n];``        ``this``.n = n;``        ``makeSet();``    ``}` `    ``// Creates n sets with single item in each``    ``public` `void` `makeSet()``    ``{``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``// Initially, all elements are in``            ``// their own set.``            ``parent[i] = i;``        ``}``    ``}` `    ``// Returns representative of x's set``    ``public` `int` `find(``int` `x)``    ``{``        ``// Finds the representative of the set``        ``// that x is an element of``        ``if` `(parent[x] != x)``        ``{``            ` `            ``// if x is not the parent of itself``            ``// Then x is not the representative of``            ``// his set,``            ``parent[x] = find(parent[x]);` `            ``// so we recursively call Find on its parent``            ``// and move i's node directly under the``            ``// representative of this set``        ``}``        ``return` `parent[x];``    ``}` `    ``// Unites the set that includes x and``    ``// the set that includes x``    ``public` `void` `union(``int` `x, ``int` `y)``    ``{``        ``// Find representatives of two sets``        ``int` `xRoot = find(x), yRoot = find(y);` `        ``// Elements are in the same set,``        ``// no need to unite anything.``        ``if` `(xRoot == yRoot)``            ``return``;` `        ``// If x's rank is less than y's rank``        ``if` `(rank[xRoot] < rank[yRoot])` `            ``// Then move x under y so that depth``            ``// of tree remains less``            ``parent[xRoot] = yRoot;` `        ``// Else if y's rank is less than x's rank``        ``else` `if` `(rank[yRoot] < rank[xRoot])` `            ``// Then move y under x so that depth of``            ``// tree remains less``            ``parent[yRoot] = xRoot;` `        ``else` `// if ranks are the same``        ``{``            ``// Then move y under x (doesn't matter``            ``// which one goes where)``            ``parent[yRoot] = xRoot;` `            ``// And increment the result tree's``            ``// rank by 1``            ``rank[xRoot] = rank[xRoot] + 1;``        ``}``    ``}``}` `// Driver code``class` `GFG``{``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``// Let there be 5 persons with ids as``        ``// 0, 1, 2, 3 and 4``        ``int` `n = 5;``        ``DisjointUnionSets dus =``                ``new` `DisjointUnionSets(n);` `        ``// 0 is a friend of 2``        ``dus.union(0, 2);` `        ``// 4 is a friend of 2``        ``dus.union(4, 2);` `        ``// 3 is a friend of 1``        ``dus.union(3, 1);` `        ``// Check if 4 is a friend of 0``        ``if` `(dus.find(4) == dus.find(0))``            ``Console.WriteLine(``"Yes"``);``        ``else``            ``Console.WriteLine(``"No"``);` `        ``// Check if 1 is a friend of 0``        ``if` `(dus.find(1) == dus.find(0))``            ``Console.WriteLine(``"Yes"``);``        ``else``            ``Console.WriteLine(``"No"``);``    ``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 `class DisjSet {``    ``constructor(n) {``        ``this``.rank = ``new` `Array(n);``        ``this``.parent = ``new` `Array(n);``        ``this``.n = n;``        ``this``.makeSet();``    ``}` `    ``makeSet() {``        ``for` `(let i = 0; i < ``this``.n; i++) {``            ``this``.parent[i] = i;``        ``}``    ``}` `    ``find(x) {``        ``if` `(``this``.parent[x] !== x) {``            ``this``.parent[x] = ``this``.find(``this``.parent[x]);``        ``}``        ``return` `this``.parent[x];``    ``}` `    ``Union(x, y) {``        ``let xset = ``this``.find(x);``        ``let yset = ``this``.find(y);` `        ``if` `(xset === yset) ``return``;` `        ``if` `(``this``.rank[xset] < ``this``.rank[yset]) {``            ``this``.parent[xset] = yset;``        ``} ``else` `if` `(``this``.rank[xset] > ``this``.rank[yset]) {``            ``this``.parent[yset] = xset;``        ``} ``else` `{``            ``this``.parent[yset] = xset;``            ``this``.rank[xset] = ``this``.rank[xset] + 1;``        ``}``    ``}``}` `// usage example``let obj = ``new` `DisjSet(5);``obj.Union(0, 2);``obj.Union(4, 2);``obj.Union(3, 1);` `if` `(obj.find(4) === obj.find(0)) {``  ``console.log(``"Yes"``);``} ``else` `{``  ``console.log(``"No"``);``}``if` `(obj.find(1) === obj.find(0)) {``  ``console.log(``"Yes"``);``} ``else` `{``  ``console.log(``"No"``);``}`

Output

```Yes
No

```

Time complexity: O(n) for creating n single item sets . The two techniques -path compression with the union by rank/size, the time complexity will reach nearly constant time. It turns out, that the final amortized time complexity is O(α(n)), where α(n) is the inverse Ackermann function, which grows very steadily (it does not even exceed for n<10600  approximately).

Space complexity: O(n) because we need to store n elements in the Disjoint Set Data Structure.